# Question 47c93

Jan 13, 2016

Here's what I got.

#### Explanation:

Your tool of choice for this problem will be the Arrhenius equation, which establishes a relationship between the rate constant of a reaction, $k$, and the temperature at which it occurs, $T$

color(blue)(k = A * "exp"(-E_a/(RT))" " " "color(purple)((1))" ", where

$A$ - a pre-exponential factor
${E}_{a}$ - the activation energy of the reaction
$R$ - the universal gas constant, in this context given as 8.3145"J"/("mol K")

Now, the Arrhenius parameters are $A$ and ${E}_{a}$.

Since you're basically dealing with one equation and two unknowns, you will have to use the information given to find a way to eliminate one of the two Arrhenius parameters first.

You can do that by writing the Arrhenius equation in non-exponential form

$\ln \left(k\right) = \ln \left[A \cdot \text{exp} \left(- {E}_{a} / \left(R T\right)\right)\right]$

This is equivalent to

$\ln \left(k\right) = \ln \left(A\right) + \ln \left[\text{exp} \left(- {E}_{a} / \left(R T\right)\right)\right]$

$\ln \left(k\right) = \ln \left(A\right) - {E}_{a} / \left(R T\right)$

Now, notice that the problem provides you with information about the value of the rate constant at two different temperatures

${T}_{1} = 273.15 + 24 = \text{297.15 K}$

${T}_{2} = 273.15 + 37 = \text{310.15 K}$

This means that you can write

$\ln \left({k}_{1}\right) = \ln \left(A\right) - {E}_{a} / \left(R \cdot {T}_{1}\right) \to$ for ${k}_{1}$ and ${T}_{1}$

$\ln \left({k}_{2}\right) = \ln \left(A\right) - {E}_{a} / \left(R \cdot {T}_{2}\right) \to$ for ${k}_{2}$ and ${T}_{2}$

Notice that you can get rid of the $\ln \left(A\right)$ term by subtracting the second equation for the first one.

$\ln \left({k}_{1}\right) - \ln \left({k}_{2}\right) = \textcolor{red}{\cancel{\textcolor{b l a c k}{\ln \left(A\right)}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{\ln \left(A\right)}}} - {E}_{a} / R \cdot \left(\frac{1}{T} _ 1 - \frac{1}{T} _ 2\right)$

This will get you

$\textcolor{b l u e}{\ln \left({k}_{1} / {k}_{2}\right) = - {E}_{a} / R \cdot \left(\frac{1}{T} _ 1 - \frac{1}{T} _ 2\right)}$

You can now find the activation energy of the reaction - keep in mind that activation energy must carry a positive sign, since it expresses energy required

${E}_{a} = - \frac{R \cdot \ln \left({k}_{1} / {k}_{2}\right)}{\frac{1}{T} _ 1 - \frac{1}{T} _ 2}$

E_a = - (8.3145 "J"/("mol" * color(red)(cancel(color(black)("K")))) * ln( (1.70 * 10^(-2) color(red)(cancel(color(black)("L mol"^(-1)"s"^(-1)))))/(2.01 * 10^(-2) color(red)(cancel(color(black)("L mol"^(-1)"s"^(-1)))))))/((1/297.15 - 1/310.15) 1/color(red)(cancel(color(black)("K"))))

${E}_{a} = \text{9873.5 J/mol}$

To get the value of the pre-exponential factor, pick a value for $k$ and $T$ and plug in the activation energy into equation $\textcolor{p u r p \le}{\left(1\right)}$.

For ${k}_{1}$ and ${T}_{1}$, you will get

${k}_{1} = A \cdot \text{exp} \left(- {E}_{a} / \left(R {T}_{1}\right)\right)$

This will get you

$A = {k}_{1} / \left(\text{exp} \left(- {E}_{a} / \left(R T\right)\right)\right)$

Notice that since the exponential has to be unitless, $A$ will have the same units as $k$. Plug in your values to get

A = (1.70 * 10^(-2)"L mol"^(-1)"s"^(-1))/("exp"(- (9873.5 color(red)(cancel(color(black)("J")))/color(red)(cancel(color(black)("mol"))))/(8.3145color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 297.15color(red)(cancel(color(black)("K"))))))#

$A = {\text{0.92475 L mol"^(-1)"s}}^{- 1}$

I'll leave the answers rounded to three sig figs.

${E}_{a} = \textcolor{g r e e n}{\text{9.87 kJ/mol}} \to$ expressed in kilojoules per mole

$A = \textcolor{g r e e n}{{\text{0.925 L mol"^(-1)"s}}^{- 1}}$